import gmpy2 from Crypto.Util.number import * import hashlib p = getPrime(1024) q = getPrime(1024) p1 = getPrime(1024) q1 = getPrime(1024) n = p * q n1 = p1 * q1 e = 0x10001 phi = (p - 1) * (q - 1) phi1 = (p1 - 1) * (q1 - 1) def chatEncrypt(pepole, message): if pepole == 'Tom': sender = hashlib.sha256(str(pepole).encode()).hexdigest() sign = hashlib.sha256(str(message).encode()).hexdigest() encryptMessage = hex(pow(bytes_to_long(message), e, n)) return sender, encryptMessage, sign elif pepole == 'Jerry': sender = hashlib.sha256(str(pepole).encode()).hexdigest() sign = hashlib.sha256(str(message).encode()).hexdigest() encryptMessage = hex(pow(bytes_to_long(message), e + 2, n)) return sender, encryptMessage, sign chatLog = open("Chatting.log", "rb") encryptedChatLog = open("ChatEncrypted.log", "w+") log = [] for line in chatLog: log.append(line) chatLog.close() encryptedChatLog.write(hex(n)+'\n') encryptedChatLog.write(hex(n1)+'\n') encryptedChatLog.write(hex(phi1)+'\n') for i in range(int(len(log)/2)): if log[i*2] == b'Tom\r\n': encryptedChatLog.write(str(chatEncrypt('Tom', log[i*2+1]))+'\n') print(chatEncrypt('Tom', log[i*2+1])) elif log[i*2] == b'Jerry\r\n': encryptedChatLog.write(str(chatEncrypt('Jerry', log[i*2+1]))+'\n') print(chatEncrypt('Jerry', log[i*2+1])) encryptedChatLog.close()解密脚本

帮我解释一下下面的代码from Crypto.Util.number import * nbit = 512 flag='****************************' p=getPrime(nbit) q=getPrime(nbit) e=65537 n=pq m= bytes_to_long(bytes(flag.encode())) P = pow(m,p,n) Q = pow(m,q,n) N=PQ phi_N=(P-1)*(Q-1) d=inverse(e,phi_N) dP=d%(P-1) print('n = ',n) print('N = ',N) print('dP = ',dP)

As an AI language model, I understand that the term "null" typically refers to a lack of value or absence of any meaningful data. It is often used in programming languages to represent the absence of ...

从以下代码中找到被藏起来的e，提示：在代码注释中。from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(25) e = # Hidden q = getPrime(25) n = p * q m = bytes_to_long(flag.strip(b"npuctf{").strip(b"}")) c = pow(m, e, n) print(c) print(pow(2, e, n)) print(pow(4, e, n)) print(pow(8, e, n)) ''' 169169912654178 128509160179202 518818742414340 358553002064450

from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(25) # Hidden e = 65537 q = getPrime(25) n = p * q m = bytes_to_long(flag.strip(b"npuctf{").strip(b"}")) c = ...

from Crypto.Util.number import * import gmpy2 import random def get_extend(): while True: d = random.randint(2680,2730) if isPrime(d): e = gmpy2.invert(d,lcm) return e flag = b"flag{xxxxxxxxxxx}" m = bytes_to_long(flag) e = 65537 p = getPrime(1024) q = getPrime(1024) lcm = gmpy2.lcm(p-1,q-1) n = p*q c = pow(m,e,n) e1 = get_extend() e2 = get_extend() print("e1 =",e1) print("e2 =",e2) print("c =",c) print("n =",n)

首先，它导入了一些必要的库，包括Crypto.Util.number、gmpy2和random。然后定义了一个函数get_extend()，它用于生成一个满足特定条件的随机数d。该函数使用random.randint()生成一个位于2^680和2^730之间...

from Crypto.Util.number import * import gmpy2 import random def get_extend(): while True: d = random.randint(2680,2730) if isPrime(d): e = gmpy2.invert(d,lcm) return e flag = b"flag{xxxxxxxxxxx}" m = bytes_to_long(flag) e = 65537 p = getPrime(1024) q = getPrime(1024) lcm = gmpy2.lcm(p-1,q-1) n = p*q c = pow(m,e,n) e1 = get_extend() e2 = get_extend() print("e1 =",e1) print("e2 =",e2) print("c =",c) print("n =",n) 这题的解题思路是什么

首先，它导入了一些必要的库，包括Crypto.Util.number、gmpy2和random。然后定义了一个函数get_extend()，它用于生成一个满足特定条件的随机数d。该函数使用random.randint()生成一个位于2680和2730之间的...

from Crypto.Util.number import * import gmpy2 def get_params(): a,b = [getPrime(128) for _ in range(2)] a,b = [bin(i)[2:].zfill(128) for i in [a,b]] p = int((a + b),2) q = int((b + a),2) a, b = int(a, 2), int(b, 2) return gmpy2.next_prime(p),gmpy2.next_prime(q),a,b flag = b'flag{******}' m = bytes_to_long(flag) p,q,a,b = get_params() n = p * q stream = [getPrime(128)] for i in range(114): num = (a * stream[-1] + b) % p stream.append(num) e = gmpy2.next_prime(stream[1] * stream[14] + stream[51] * stream[4]) c = pow(m,e,n) print(f'n = {n}') print(f'c = {c}') # n = 6983103504667653623549839927036982238702989760981440695828662474048376735590137382537887815211794982723527122829225808557810219686458488017050037179654123 # c = 6796222297027759937096057530534388748539384795006307524526800462652134886498275107011127280721467103062302906430290388832305421496018525958278394725734527 写个脚本解出flag

from Crypto.Util import long_to_bytes import gmpy2 def get_params(): a, b = [getPrime(128) for _ in range(2)] a, b = [bin(i)[2:].zfill(128) for i in [a, b]] p = int((a + b), 2) q = int((b + a), 2...

from Crypto.Util.number import * from random import randint from secret import flag def TrickPrime(bits): p = getPrime(bits) q = getPrime(bits) cut = randint(1,256) temp = p*q print('clown =',temp) game = (p&(2**bits-1)) >>cut<<cut #p高位需要给出 print("trick =",game) return p,q def CrazyPrime(nbits): p = getPrime(nbits) q = getPrime(nbits) r = getPrime(nbits) n = p * q * r print("n =", n) m = getPrime(256) P = pow(m, p, n) Q = pow(m, q, n) R = pow(m, r, n) print("P =", P) print("Q =", Q) print("R =", R) return m P,Q = TrickPrime(512) R = CrazyPrime(512) N =PQR phi = (P-1)(Q-1)(R-1) e = 65537 d = inverse(e,phi) m = bytes_to_long(flag) c = pow(m,e,N) print("c = ",c)

CrazyPrime生成三个质数p、q和r，然后返回一个随机生成的数m，以及三个数P、Q和R，其中P、Q和R分别是$m^p \bmod n$、$m^q \bmod n$和$m^r \bmod n$，其中$n=pqr$。这段代码的加密过程是将...

from Crypto.Util.number import * from gmpy2 import * from secret import flag flag = flag.strip(b'HNZUCTF{').strip(b'}') class textbookRSA: def init(self, p, q, e): self.p = p self.q = q self.N = self.p * self.q self.e = e self.d = invert(self.e, (self.p - 1) * (self.q - 1)) def encrypt(self, m): assert m < self.N return pow(m, self.e, self.N) def decrypt(self, c): return pow(c, self.d, self.N) def vulnerability(self): vul = [pow(2, self.e, self.N), pow(4, self.e, self.N), pow(8, self.e, self.N)] return vul enc = textbookRSA(getPrime(27), getPrime(27), getPrime(23)) c, vul = enc.encrypt(bytes_to_long(flag)), enc.vulnerability() print(c) print(vul) ''' 9219741073542293 [7881951730741780, 4599083407040344, 3939540884944030] '''

from Crypto.Util.number import * from gmpy2 import * from secret import flag flag = flag.strip(b'HNZUCTF{').strip(b'}') class textbookRSA: def __init__(self, p, q, e): self.p = p self.q = q self...

import os import signal from Crypto.Util.number import * flag = os.environb.get(b"FLAG", b"dummmmy{test_test_test}") def main(): p = getPrime(128) q = getPrime(128) n = p * q N = pow(p, 2) + pow(q, 2) print("Let's factoring !") print("N:", N) p = int(input("p: ")) q = int(input("q: ")) if isPrime(p) and isPrime(q) and n == p * q: print("yey!") print("Here you are") print(flag) else: print("omg") def timeout(signum, frame): print("Timed out...") signal.alarm(0) exit(0) if name == "main": signal.signal(signal.SIGALRM, timeout) signal.alarm(30) main() signal.alarm(0)

它生成两个128位的随机质数p和q，并计算出它们的乘积n和平方和N。然后，程序要求你输入p和q，并检查它们是否是质数并且乘积等于n。如果条件满足，它将打印出FLAG的值。如果条件不满足，则打印"omg"。此外，程序设置...

#encoding:utf-8 from Crypto.Util.number import * from gmpy2 import * from flag import flag def gen(MaxBits, Bits): while True: p = getPrime(Bits) q = getPrime(Bits) N = (p**r)*q if len(bin(N)) -2 == MaxBits: break phi = (p**(r-1))(p-1)(q-1) idx = (r(r-1)) / ((r+1)(r+1)) delta = int(pow(mpz(N), idx)) while True: d1 = getPrime(int(2048idx)//2) d2 = getPrime(int(2048idx)//2) if abs(d1-d2) < delta: e1 = invert(d1, phi) e2 = invert(d2, phi) break e = 0x10001 return N, e, e1, e2 Bits = 256 MaxBits = 2048 N, e, e1, e2 = gen(MaxBits, Bits) M = bytes_to_long(flag) C = powmod(M, e, N) print(f"N={N}\nC={C}\ne={e}\ne1={e1}\ne2={e2}\n") ''' N=26989781630503676259502221325791347584607522857769579575297691973258919576768826427059198152035415835627885162613470528107575781277590981314410130242259476764500731263549070841939946410404214950861916808234008589966849302830389937977667872854316531408288338541977868568209278283760692866116947597445559763998608870359453835826711179703215320653445704522573070650642347871171425399227090705774976383452533375854187754721093890020986550939103071021619840797519979671188117673303672023522910200606134989916541289908538417562640981839074992935652363458747488201289997240226553340491203815779083605965873519144351105635977 C=15608493359172313429111250362547316415137342033261379619116685637094829328864086722267534755459655689598026363165606700718051739433022581810982230521098576597484850535770518552787220173105513426779515790426303985414120033452747683669501078476628404455341179818932159581239994489678323564587149645006231756392148052557984581049067156468083162932334692086321511063682574943502393749684556026493316348892705114791740287823927634401828970155725090197482067045119003108806888768161101755244340832271562849138340706213702438667804460812804485276133545408754720942940596865774516864097546006862891145251661268265204662316437 e=65537 e1=8334176273377687778925968652923982846998724107624538105654894737480608040787164942908664678429487595866375466955578536932646638608374859799560790357357355475153852315429988251406716837806949387421402107779526648346112857245251481791000156326311794515247012084479404963628187413781724893173183595037984078029706687141452980915897613598715166764006079337996939237831127877822777298891345240992224457502307777453813403723860370336259768714433691700008761598135158249554720239480856332237245140606893060889458298812027643186014638882487288529484407249417947342798261233371859439003556025622531286607093086262182961900221 e2=22291783101991466901669802811072286361463259096412523019927956845014956726984633944311563809077545336731345629003968417408385538540199052480763352937138063001691494078141034164060073208592072783644252721127901996835233091410441838546235477819239598146496144359952946239328842198897348830164467799618269341456666825968971193729838026760012332020223490546511437879465268118749332615890600046622926159177680882780495663448654527562370133394251859961739946007037825763819500955365636946510343942994301809125029616066868596044885547005547390446468651797783520279531291808102209463733268922901056842903640261702268483580079 ''' 请运用copperSmith定理解出M

from Crypto.Util.number import long_to_bytes from gmpy2 import invert, iroot, mpz # 给定参数 N = ...

#encoding:utf-8 from Crypto.Util.number import * from gmpy2 import * from flag import flag def gen(MaxBits, Bits, r): k =8 while True: p = getPrime(Bits) q = getPrime(Bits) N = (p**r)*q if len(bin(N)) -2 == MaxBits: break phi = (p**(r-1))(p-1)(q-1) idx = (r(r-1)) / ((r+1)(r+1)) delta = int(pow(mpz(N), idx)) while True: d1 = getPrime(int(MaxBitsidx)//2) d2 = getPrime(int(MaxBitsidx)//2) if abs(d1-d2) < delta: e1 = invert(d1, phi) e2 = invert(d2, phi) break e = 0x10001 return N, e, e1, e2 r = 7 Bits = 256 MaxBits = 2048 N, e, e1, e2 = gen(MaxBits, Bits, r) M = bytes_to_long(flag) C = powmod(M, e, N) print(f"N={N}\nC={C}\ne={e}\ne1={e1}\ne2={e2}\n") ''' N=26989781630503676259502221325791347584607522857769579575297691973258919576768826427059198152035415835627885162613470528107575781277590981314410130242259476764500731263549070841939946410404214950861916808234008589966849302830389937977667872854316531408288338541977868568209278283760692866116947597445559763998608870359453835826711179703215320653445704522573070650642347871171425399227090705774976383452533375854187754721093890020986550939103071021619840797519979671188117673303672023522910200606134989916541289908538417562640981839074992935652363458747488201289997240226553340491203815779083605965873519144351105635977 C=15608493359172313429111250362547316415137342033261379619116685637094829328864086722267534755459655689598026363165606700718051739433022581810982230521098576597484850535770518552787220173105513426779515790426303985414120033452747683669501078476628404455341179818932159581239994489678323564587149645006231756392148052557984581049067156468083162932334692086321511063682574943502393749684556026493316348892705114791740287823927634401828970155725090197482067045119003108806888768161101755244340832271562849138340706213702438667804460812804485276133545408754720942940596865774516864097546006862891145251661268265204662316437 e=65537 e1=8334176273377687778925968652923982846998724107624538105654894737480608040787164942908664678429487595866375466955578536932646638608374859799560790357357355475153852315429988251406716837806949387421402107779526648346112857245251481791000156326311794515247012084479404963628187413781724893173183595037984078029706687141452980915897613598715166764006079337996939237831127877822777298891345240992224457502307777453813403723860370336259768714433691700008761598135158249554720239480856332237245140606893060889458298812027643186014638882487288529484407249417947342798261233371859439003556025622531286607093086262182961900221 e2=22291783101991466901669802811072286361463259096412523019927956845014956726984633944311563809077545336731345629003968417408385538540199052480763352937138063001691494078141034164060073208592072783644252721127901996835233091410441838546235477819239598146496144359952946239328842198897348830164467799618269341456666825968971193729838026760012332020223490546511437879465268118749332615890600046622926159177680882780495663448654527562370133394251859961739946007037825763819500955365636946510343942994301809125029616066868596044885547005547390446468651797783520279531291808102209463733268922901056842903640261702268483580079 ''' 使用coopersmith定理解出m

from Crypto.Util.number import long_to_bytes from gmpy2 import invert, iroot, mpz # 已知参数 N = ...

from Crypto.Util.number import * from secret import flag, x, y def keygen(nbit): p, q = [getPrime(nbit) for _ in range(2)] return (p, q) p, q = keygen(1024) n = p * q t = len(flag)//2 part1 = bytes_to_long(flag[:t]) part2 = bytes_to_long(flag[t:]) D = 1117 x = y = assert x2 - D * y2 == 1 enc1 = pow(233 * n 2 + 1, part1, n 3) enc2 = pow(y * n + 1, part2, n ** 3) print(f'n = {n}') print(f'enc1 = {enc1}') print(f'enc2 = {enc2}')请解密出flag

p, q = [getPrime(nbit) for _ in range(2)] return (p, q) p, q = keygen(1024) n = p * q 接下来，我们需要求出 $y$，满足 $x^2 - Dy^2 = 1$。这可以使用 Sympy 库来求解。 python from sympy import ...

from Crypto.Util.number import * from Crypto.PublicKey import RSA from secret import s, FLAG def gen_prime(s): while True: r = getPrime(s) R = [r] t = int(5 * s / 2) + 1 for i in range(0, t): R.append(r + getRandomRange(0, 4 * s ** 2)) p = reduce(lambda a, b: a * b, R, 2) + 1 if isPrime(p): if len(bin(p)[2:]) == 1024: return p while True: p = gen_prime(s) q = gen_prime(s) n = p * q e = 65537 d = inverse(e, (p-1)(q-1)) if len(bin(n)[2:]) == 2048: break msg = FLAG key = RSA.construct((long(n), long(e), long(d), long(p), long(p))) for _ in xrange(s): enc = key.encrypt(msg, 0)[0] msg = enc print(key.publickey().exportKey()) print('-' 76) print(enc.encode('base64')) print ('-' * 76)

from Crypto.Util.number import getPrime, inverse, getRandomRange from Crypto.PublicKey import RSA import base64 2. 修改 long 函数为 int： python key = RSA.construct((int(n), int(e), ...

import os import ecdsa import hashlib from Crypto.Util.number import * from Crypto.Util.strxor import strxor as xor import secret p = getPrime(256) gen = lambda: p + getPrime(16) pad = lambda m: m + os.urandom(32 - len(m) % 32) key = os.urandom(30) sk = ecdsa.SigningKey.from_secret_exponent( secexp=bytes_to_long(key), curve=ecdsa.SECP256k1 ) sig1 = sk.sign(data=b'This is the first message.', k=gen()).hex() sig2 = sk.sign(data=b'Here is another message.', k=gen()).hex() enc = xor(hashlib.sha512(key).digest(), pad(secret.flag)).hex() print(f"{sig1 = }\n{sig2 = }\n{enc = }") ''' sig1 = '3f4a6f288e35a4397201d246a98c1f9cfa463e67717fbbdcbd26d7fac75f875855455c2bfb355f7f593ffbe4c4bd1fc729cc129976b56905639100c8ac716b37' sig2 = '9f563b21f0ee31b2f7a1a8c6edc8ff23b63e0a9d5dd4a699ecc3164871b4982df51bb2feb4bc06c578afd21d3e6227231dd5fe1d8440f3dcd025fd3ea68f5516' enc = 'cc66d251bfa54954890c81dc1c607bae716573949f327db18aa1f4c0f420b8d29dc7e7ff9edb17b90306bd2aa753fc3fd4dafb9cc4b771cbdd79000ef05a40c0' '''

1. 生成一个 256 位的质数 p，然后再随机生成一个 16 位的质数，将两个相加得到一个 272 位的数作为 ECDSA 的生成元。 2. 随机生成一个 30 字节的密钥 key，使用 SHA-512 哈希函数对其进行哈希，得到一个 64 字节的...

from Crypto.Util.number import * from sympy import * from secret import flag def keygen(nbit): p, q = [getPrime(nbit) for _ in range(2)] return (p, q) p, q = keygen(1024) n = p * q D = 1117 x = y = symbols('x y') eq = Eq(x**2 - D*y**2, 1) sol = solve(eq, (x, y)) x, y = sol[0] factors = factorint(n) p, q = factors.keys() u1 = pow(233*n**2+1, p-1, p) u2 = pow(233*n**2+1, q-1, q) v = (u1qinverse(q, p) + u2pinverse(p, q)) % n t = len(flag)//2 enc1 = int(input("enc1:")) enc2 = int(input("enc2:")) part1 = pow(enc1, inverse((p-1)(q-1), n), n) part2 = pow(enc2, (p+1)//4 (q+1)//4, n) flag = long_to_bytes(part1) + long_to_bytes(part2) print(flag)请运行这段代码

根据 Tonelli–Shanks 算法，可以计算出 $m_2^{(p+1)/4}$ 和 $m_2^{(q+1)/4}$ 的平方根 $r_p$ 和 $r_q$，使得 $r_p^2 \equiv m_2^{(p+1)/4} \pmod{p}$，$r_q^2 \equiv m_2^{(q+1)/4} \pmod{q}$，然后使用 CRT 合并 $...

from Crypto.Util.number import getPrime, bytes_to_long from os import urandom from secret import flag def padding(m, n): assert len(m) <= n return m + urandom(n - len(m)) n = getPrime(2048) * getPrime(2048) g = n + 1 x = bytes_to_long(padding(flag, n.bit_length()//8-1)) c = pow(g, x, n*n) print('n = %d' % n) print('c = %d' % c)知道了n和c的值怎么求flag的值

As an AI language model, I don't have emotions or personal opinions. "Null" is a term used in programming to represent a value that is unknown or has no value. It can also refer to a condition of ...

#!/usr/bin/env python3 import os from Crypto.Util.number import getPrime, getRandomRange def isSquare(a, p): return pow(a, (p-1)//2, p) != p-1 class SquareRNG(object): def init(self, p, sa, sb): assert sa != 0 and sb != 0 (self.p, self.sa, self.sb) = (p, sa, sb) self.x = 0 def int(self, nbits): v, s = 0, 1 for _ in range(nbits): self.x = (self.x + 1) % p s += pow(self.sa, self.x, self.p) * pow(self.sb, self.x, self.p) s %= self.p v = (v << 1) | int(isSquare(s, self.p)) return v def bool(self): self.x = (self.x + 1) % self.p t = (pow(self.sa, self.x, self.p) + pow(self.sb, self.x, self.p)) t %= self.p return isSquare(t, self.p) p = getPrime(256) sb1 = int(input("Bob's seed 1: ")) % p sb2 = int(input("Bob's seed 2: ")) % p for _ in range(77): sa = getRandomRange(1, p) r1 = SquareRNG(p, sa, sb1) print("Random 1:", hex(r1.int(32))) r2 = SquareRNG(p, sa, sb2) print("Random 2:", hex(r2.int(32))) guess = int(input("Guess next bool [0 or 1]: ")) if guess == int(r1.bool()): print("OK!") else: print("NG...") break else: print("Congratz!") print(os.getenv("FLAG", "nek0pts{* REDACTED *}"))

from Crypto.Util.number import getPrime, getRandomRange def isSquare(a, p): return pow(a, (p-1)//2, p) != p-1 class SquareRNG(object): def __init__(self, p, sa, sb): assert sa != 0 and sb != 0 ...

i2pdbrowser：实现Firefox通过I2P安全连接的捆绑工具

资源摘要信息:"i2pdbrowser是一个包含了脚本的浏览器捆绑包，这些脚本能够下载最新的Firefox Extended Support Release（ESR）版本，并将其配置为通过I2P网络使用i2pd工具进行通信。以下是关于该资源的详细介绍，...

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